Stuck on specific argument in the proof of O'Nan-Scott-Aschbacher theorem

I'm stuck on an argument from the last paragraph in the paper "On the O'Nan-Scott theorem for finite primitive permutation groups", https://doi.org/10.1017/S144678870003216X ,by Liebeck, Praeger and Saxl, which is a proof of the O'Nan-Scott-Aschbacher theorem that classifies finite primitive permutation group. Need help. I will quote the relevant part.

In this context, $G$ is a finite primitive permutation group. $T$ is a normal subgroup of $G$ which is also a nonabelian simple group, and the conjugation action of $G$ on $T$ induce an injection of $G$ into $Aut(T)$. $\alpha$ is a point in the space, and any group written with subscript $\alpha$ mean the stabilizer subgroup of $\alpha$ in that group. The stabilizer of $\alpha$ in $G$, written as $G_{\alpha}$, is known to be a maximal subgroup of $G$. Notation: $C,N$ are centralizer and normalizer, and $[\cdot,\cdot]$ is commutator. The argument starts by assuming $T_{\alpha}=1$ and attempt to derive a contradiction. The argument is as follow:

Then $G_{\alpha}$ is soluble by the Schreier "Conjecture". Let $Q$ be a minimal normal subgroup of $G_{\alpha}$. Then $Q$ is an elementary abelian $q$-group for some prime $q$. Now $C_{T}(Q)=1$ since both $T$ and $C_{G}(Q)$ are normalized by $G_{\alpha}$ and $G_{\alpha}$ is maximal in $G$. It follows that $q$ does not divide $|T|$. Hence $Q$ normalizes a Sylow 2-subgroup $S$ of $T$. We assert that $S$ is the unique such Sylow 2-subgroup. For suppose that $Q$ normalizes $S1$, where $S1^{x}$ =S and $x\in T$. Then $Q$ and $Q^{x}$ are Sylow $q$-subgroups of $N_{TQ}(S)$, so $Q=Q^{xy}$ for some $y\in N_{T}(S)$. We have $[Q,xy]\leq Q\cap T=1$ so $xy\in C_{T}(Q)=1$. Hence $x\in N_{T}(S)$ so $S1=S$ as asserted. Thus $G_{\alpha}\leq N_{G}(Q)\leq N_{G}(S)$ and so $G_{\alpha}<G_{\alpha}S<G$ contradicting the maximality of $G_{\alpha}$.

The part I don't get is this:

It follows that $q$ does not divide $|T|$. Hence $Q$ normalizes a Sylow 2-subgroup $S$ of $T$.

I'm not sure why $q$ does not divide $|T|$ from the previous argument, and I'm not sure how, from that, we know that $Q$ normalizes a Sylow 2-subgroup of $T$.

Solution 1:

$Q$ acts on $T$ by conjugation. Every orbit of this action, except the orbit of $1$ is non-singleton because $C_T(Q)=1$. The size of every orbit divides $|Q|$, so is a power of $q$. Hence $|T|\equiv 1\pmod{q}$. Hence $q \nmid |T|$.

Also $Q$ acts on the set $P$ of Sylow $2$-subgroups of $T$ by conjugation, the number $|P|$ of them divides $|T|$ by the Sylow theorem. The size of every orbit of this action is a power of $q$. If all orbits have $>1$ elements, $|P|$, and hence $|T|$, would be divisible by $q$.

Remark I have not seen papers about finite group theory for long time (the last one was by Gorenstein and Walter) but I remember that arguments of this kind appear so often there that the authors usually do not include them in proofs.